The lifespans of lions in a particular zoo are normally distributed. The average lion lives $12.6$ years; the standard deviation is $2.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $6.8$ years.
Explanation: $12.6$ $9.7$ $15.5$ $6.8$ $18.4$ $3.9$ $21.3$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $12.6$ years. We know the standard deviation is $2.9$ years, so one standard deviation below the mean is $9.7$ years and one standard deviation above the mean is $15.5$ years. Two standard deviations below the mean is $6.8$ years and two standard deviations above the mean is $18.4$ years. Three standard deviations below the mean is $3.9$ years and three standard deviations above the mean is $21.3$ years. We are interested in the probability of a lion living less than $6.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $6.8$ years and the other half $({2.5\%})$ will live longer than $18.4$ years. The probability of a particular lion living less than $6.8$ years is ${2.5\%}$.